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What is the function of a skew sensor in an aircraft?
I was trying to understand Slat and Flap control Unit and came across the term "skew sensor". Can anyone explain it to me?

7 upvotes? Guys does this question show research effort?

Skew is essentially when the flap doesn't extend or retract straight back.
Picture a situation where you have a 1 meter wide board you're pushing down a track that's slightly wider than the board.  Your left hand is about a quarter of the way in from the left edge, and your right hand similarly in from the right edge.  As long as you push straight down the track, evenly, with both hands, the board slides along and doesn't contact the edge.  Now suppose that your right hand pushes harder than your left hand: the board will start to turn, and the edges of the board will contact the edges of the track.
The skew sensor detects that this sort of misalignment is starting to happen, and causes the flap drive to stop.  Ideally, this takes place before the flap makes physical contact with any other hardware, so nothing is damaged!
Edit:
Video of flaps being extended & retracted on the ground.  You can see that there are multiple flaps on each wing, and while normally everything moves in synch, you can envision how if one actuator stopped working, the errant flap could get turned pretty far out of alignment, make contact with other hardware, and cause some significant damage.  So before this happens, the skew sensor sends the command back to stop everything from moving.
From Boeing's Aero Magazine:

The 757-300 has incorporated a flap skew detection system. Flap skew occurs when either the inboard or outboard edge of the flap moves farther than the rest of the flap. The flap skew detection system monitors, detects, and shuts down the flaps if a skew condition occurs. The heart of this system is the flap slat accessory module (FSAM), taking the place of the flap slat electronic unit (FSEU) number 3. The new LRU was introduced because the existing configuration of the FSEU would not accommodate the addition of the flap skew circuits.
The flap skew sensors are mounted on the trailing edge of the flap jack screws (fig. 4a). The sensor (fig. 4b) is fixed to the flap track, and a target (which contains magnets) is mounted to the jackscrews. As the jackscrew begins to turn, the target also turns. This will pulse the sensor, and a signal (distance traveled) is sent to the FSAM, which compares this signal with the opposite side of the same flap. If the distances traveled by either side of the flap differ by a predetermined amount, the FSAM will shut down the flap system to help eliminate damage to the flap and surrounding structure and the possible loss of a flap in flight. The alternate flap mode will override the normal flap skew detection function, allowing maintenance personnel to move the flaps after maintenance has been performed. Maintenance personnel must verify the flap fault and also ensure that moving the flap will not cause any damage.

What's the analog of the left & right hands here? Two hydraulic actuators?

@curious_cat  Typically, two jackscrews that are driven by hydraulic motors.


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Under what circumstances do planes poop?
I had always assumed that human waste was removed from commercial aircraft while they were on the ground, some inscrutable machine coming by to suck the material out and carry it away.
Although that may happen, it appears that planes do have the ability to poop and they use that ability as one woman found out yesterday after a plane pooped on her house.
Apparently commercial carriers euphemistically call the poop "blue ice".
Under what circumstances do planes eject blue ice while airborne rather than dispose of it when on the ground?

Note that the alleged duplicate question does not answer the question correctly so I have posted a correct answer here below.

It is almost always the result of a leakage in the treatment system, no aircraft that I know of has the ability to purposely discharge the waste system.

@RonBeyer Looks like you have an answer, why limit that to a comment?

@KorvinStarmast It's the "that I know of" that keeps me from that, apparently there were some 727's that had this ability. I don't know of any modern aircraft that has this "ability", but I could be mistaken, and haven't really looked into it enough to constitute a factual answer.

@RonBeyer There was a thread on PPRuNe recently that had some discussion of this particular issue.  I'll see if I can find it.

This schematic of the B737NG water/waste system can be found in this question. Waste tank collects and store toilet waste, including flush water, and is emptied only on the ground. Vacuum is created by a pump protected by a water separator and a filter. There could be a malfunction at this level.

@RonBeyer I thought that was a malfunction with the 727, not an intentional ability?

I thought airlines pooped over the ocean.  Maybe the pilot didn't want to pollute the ocean.  I wonder if the poop system on airliners is automatic or manually done.

@user6035379 You thought wrongly. There is no "poop system" in the sense that you seem to be using the phase.

This question is about blue ice, not pee.

After some research I discovered that "blue ice" jetsam from commercial planes are not uncommon and in several instances people have even been struck by poo bombs, although apparently noone has been killed yet.
Normally, the septic tank on a plane is evacuated on the ground by a special truck after the plane has landed. However, the septic tanks have an overflow valve. If, for any reason, the septic tank becomes full and pressurized, then it will start to overflow outside of the plane through the waste drain valve assembly (shown in diagram below). This overflow can accumulate into a ball of ice due to the freezing cold temperatures at high altitudes. The ice then falls off either when it gets too big or when the temperatures warm it as the plane nears the ground.

This answer would be more authoritative if you were able to provide documentation for your statements regarding the overflow vent and the tank system. Such documentation should define the type of aircraft discussed as well include such information as the system's operating description and schematic.

I studied the my maintenance manual for one of the older B737 models and it does not have an overflow valve from the waste water system. There is, however, an overflow valve from the potable water system which would dump overflow water overboard. In the absence of any evidence to the contrary I am going to suggest that you are confusing the two systems.

@JonathanWalters Out of curiosity, what could cause the potable water system to overflow during flight? It seems like that would only be possible when filling it on the ground.

@reirab The overflow drain is only opened when the combination fill/overflow valve is opened. Overflow would only occur from over-servicing on the ground. In normal flight operation the potable water tank is pressurized by bleed air with the only open outlet being the plumbing to the faucets. Theoretically, a valve failure accompanied by some combination of a series of negative and positive G loading or excessive tank pressurization might lead to minimal dumping overboard in flight, but such a scenario would be a highly unlikely, if possible at all.

@JonathanWalters If the tank is full, it will overflow. It is that simple.

@TylerDurden Where will it overflow to? The lav floor? Where will it overflow from?

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Forces that wing attach fittings must transfer to fuselage?
I just saw the naked wing (it didn't have its wing fabric) of a Weedhopper ultralight and wondered the types of forces that the wing lift struts must transfer to the fuselage, apart from the apparent lift force?
Secondly, in a wire braced wing like the Weedhopper, do lift struts carry the lift load on the span between their attach point and the fuselage or do they carry the lift on the span from the attachment point to the wing tip?

I made some assumptions for your second question, please have a look at it in case that is not what you meant.

I've made an insanely oversimplified sketch to show the general loads on something like a weedhopper.  All of the attachments are pin joints that can't resist rotation about the pin, and therefore take shear loads only.
It's a simple pin jointed triangle, complicated by the fact that the main element is a long beam extended beyond the triangle, loaded across its axis.  Most of the weight of the airplane is on the main lift struts, with mostly a compression force on the root fittings but with some up or down shear as well, depending.
The shear loads on the root fittings depend a lot on where the main lift strut picks up the wing, having an upward/inboard component, downward/inboard component, or neutral/inboard component depending on where the lift strut is attached along the spar beam.
Drag and anti-drag loads on the wing, and pitching moments also add or subtract from the shear loads caused by the lift force on the root fittings. Working it all out gets pretty complicated.
The root fitting loads are usually lower than the strut attach loads out along the wing and you'll often see surprisingly small bolts attaching the wing to the fuselage on high wing aircraft.  On little puddle jumper planes like Aeroncas, you'll see fairly beefy 3/16" or 1/2" bolts at the wing to lift strut attach, but itty bitty quarter inch bolts joining the wing to the fuselage.
On a Twin Otter, with strut attachments way inboard at the nacelles, the root shear loads are more substantial and are inboard and down.
On the Weedhopper my guess is the lifting force transmits an upward and inboard shear load to the root bolts and this force varies with drag loads and pitching moments.

For your first question, it depends on the way they are attached, as is often the case in structures. I haven't seen this model up close to know their fitting type, but:

If they are fixed by a joint that allows them to rotate (the easiest and cheapest solution) then they can only carry one kind of load, that being tensile (or compressive) stresses. this would mean they can only transmit force along their own axis.
If they are rigidly jointed (cannot rotate at the joint) then they also carry some torque, although how much is a matter for the structural engineers. My guess is not much as the attachment points already offer a large moment arm.

For your second question, I believe you mean to ask for what part of the overall lift are the lift struts responsible for. That I cannot readily answer without some calculations I don't have time for right now, but it is not as straightforward as you imply. The Wing-Fuselage-Strut structure is hyperstatic (statically indeterminate or over-constrained) so it is a bit too long to napkin-math.

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What are the relative contributions of the body and wings to missile lift?
Slender bodies like arrow shafts, airliner fuselages, and submarine hulls are all capable of generating lift, albeit much less efficiently than a high aspect ratio wing. They also generate little drag, which is their main advantage. I would like to know how their lifting properties compare to that of a wing; to this end I would like to refer to a use case where the slender body makes up most of the airframe: guided missiles.

Consider a missile with a long slender body and short lifting surfaces, with a "conventional" air to air missile aspect ratio like that of the R-27 or AIM-120. The exact wing and control surface layout is not relevant, only the form factor. This means we should purposefully exclude munitions equipped with a moderate aspect ratio wing such as the TLAM or JSOW: I am interested in low aspect ratio BVR designs:

To avoid complications due to the engine or absence thereof, consider it flying in the coasting phase. The lift needed to stay aloft or even climb is then provided by a combination of the wings and the slender main body (where the guidance systems, warhead and sometimes engine are housed).
The question is, what is the proportion of lift coming from the wings compared to that of the fuselage?
NOTES:

A good answer should use a real or realistic design to base their calculations on.
I have purposefully left the choice of design open, so people can pick one they are familiar with or have data on. All that matters to me is that the general shape must be similar to that of modern AAMs.
Feel free to elaborate on how the lift proportion changes at different flight conditions (subsonic vs supersonic); it is quite clear that higher speeds lead to smaller wings.
I am aware that short range WVR missiles like the R-60, R-73 or AIM-9 are designed to reach their target before engine cutoff or shortly after; this has no bearing on the question.
This answer below presents a good extreme where most of the lift has to come from the body, because the only fins are too far away from the center of mass. In such cases, where the fins act as a stabilizer, their lifting force can still be calculated and compared to the total lift required, if we can assume the location of the CG.

The missiles you refer to do not typically operate "in the coasting phase" - they are supposed to strike before they "run out of gas".

@MikeBrockington if you have a source I would really like to see it. To my knowledge burn time of the AIM-120 is about 6 seconds, which would not account for most of it's range (source). That being said, assuming engine cut off is a simplification that should not impact the calculations by more than $Sin (\theta)$ with $\theta \ll 1$ and can be disregarded for a first approximation.

@MikeBrockington Only very short-range missiles have the luxury of hitting before the gas runs out. The exact properties of the solid fuel and the fuel-fraction/aerodynamics of the missiles may not be openly published, but are simple enough that reasonably-detailed simulations can show they won't get anywhere near the designed range if they could only make the kill with the engine active.

The fins (I won't call them wings) are axially symmetric, so they're more for steering than for countering gravity.  Even so, they likely produce significant lift, as even a small wing at high speed does (F-104, X-15, etc.).

@CamilleGoudeseune at my university we used the term wings to distinguish them from moveable control surfaces and traditional fins (fixed surfaces at the rear whose main purpose is providing stability). But whether they can be rightfully called wings (in this case) is part of my question :)

Seems like missiles can't produce horizontal flight without thrust: https://www.quora.com/How-do-missiles-fly-horizontally

@Jimmy neither can a glider... your point? The fact is that missiles can maneuver in the coasting phase, so they must be able to generate lift at the expense of increased drag, and trade kinetic energy for height like any other aircraft.

It is actually important to realize that "hunter" missiles like the Sidewinder have central "fins" to destabilize the missile so it can turn and close with its target.  Also, greater speed and a lower weight to surface area ratio works in favor of "bodylifters". But notice subsonic cruisers do need a wing.  Supersonic cruisers may make clever use of their "bow wave" to produce lift, a technique also used by the much larger B-70.  Perhaps this needs to be accounted for in supersonic lifting formulas.

GMC, Funny, gliders are "ballistic" too, until they generate enough lift to glide!  Check out Apollo capsule data (a must for plasma lovers).  Slightly offsetting the CG enough to make the capsule "glide" helped solve the re-entry at 25,000 mph issue.

@AEhere I mean to say that missiles cannot perform 1G flight without thrust. Gliders perform 1G flight almost 100% of the time. Missiles, on the other hand, are almost ballistic once depleted of fuel. The fins generate aerodynamic forces but only for stabilization and (in some designs) course corrections.

Maybe the question could be better thought of as:  when the missile is turning, what's contributing most to the turning force?  "Lift", if thought of as the force always opposing gravity, is basically entirely from the rocket motor, but at some point any guided missile worth its salt is pulling turns of ~30G.  Surely the control surfaces start the turn, but then is the body contributing when the AoA is high?

What is the proportion of lift coming from the wings compared to that of the fuselage?

It can be almost zero.  The ASRAAM's only fins or wings or lifting surfaces or whatever you call 'ems are at its very tail.  Any "lift" they impart, that far from the missile's center of mass, would pitch (or yaw) the missile instead of counteracting gravity.  So almost all the lift must come from the body having a slight nose-up attitude.

Another missile with a similar planform is the (ground vehicle launched) LOSAT.

Yet the missile remains in straight flight, so we could, from its characteristics, quantify the amount of force generated by the fins and the body, and compare them. In fact, in that configuration, the fins are lifting, or the missile would pitch up.

Also, +1 for the LOSAT, not a usual example, but one I remember from the books I had as a kid.

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Did these weather conditions require the airline to cancel a flight?
[I do not believe this question duplicates a previous question, which involves different conditions and circumstances]
I'm interested in the airline's response to a query of mine about why a flight was cancelled due to weather when destination airport had visibility of at least 12 SM. I asked them and the airline wrote back, attaching relevant METAR and TAF reports.
The destination airport was Victoria International (CYYJ) which is not category III (maybe cat II?). All METAR reports indicated 12SM visibility with the lowest ceiling FEW170. The airline acknowledges that the actual conditions were clear. The flight was scheduled to depart 0716 ZULU, 33 min. duration, 2 alternates with CAT III available at 16 min (Vancouver, CYVR) and 33 min (Seattle, KSEA) flying time.
The aircraft was a Dash 8. A competing airline flew the same route with a CRJ 15 minutes before the scheduled departure without incident. This was the only cancellation at YYJ that evening (there were perhaps a dozen arrivals, mostly Dash 8s or Q400s but there was an A320 and at least one 737).
First question
The airline claims a TEMPO condition of 1/4SM made it illegal to dispatch. But the TAF at 335Z said TEMPO 1 1/2SM BR SCT002, not 1/4SM. The bases prediction was P6SM SCT250 but there was also a PROB40 for 1/4SM and fog (FG).
Is a PROB40 for fog and 1/4 SM visibility enough to cancel? Is the airline required to cancel at that point? The 335Z TAF also had RMK next forecast by 0600Z, 1 hour before departure.
Second question
The new TAF issued at 538Z did not have TEMPO but had PROB30 (I can't find PROB30 in my TAF/METAR definition list, as PROB40 is supposed to be 30-50%, right?) for 1/4SM and fog (FG) again with what appears to be a main prediction of P6SM FEW150 SCT250. Was this enough to reasonably cancel?  Was the airline legally required to cancel on this basis?

Update: a poster asked me if I could post the full set of TAF and METAR reports, as received from the airline. Here they are (assuming no typos on my part because I received these hard-copy not electronic):
Scheduled flight departure from SEA  Zulu 7:16  Flight duration: 33 min. Arrival YYJ
METARS for YYJ:

0700Z  27004KT 12SM FEW170 BKN250 01/01 A3043
  0600Z  27004KT 12SM SCT250 M01/M01 A3041
  0500z  27004KT  15SM FEW250 00/M00 A3041

TAFs for YYJ:

CYYJ 1005338Z 1006/1106 VR03KT  P6SM FEW150 SCT250
  PROB30 1006/1016 1/4SDM FG VV002 FM101800 29008KT P6SM FEW004 SCT100
  FM102000 35005KT P6SM SKC RMK NXT FCST BY 101200Z
CYYJ 100335Z 1003/1024  VRB03KT P6SM SCT250
  TEMPO 1003/10-18 1 1/2SM  BR SCT002
  PROB40 1005/1016 1/4 SM FG VV002 FM101800 29008KT P6SM
  FEW004 RMK NXT FCST BY 100600Z

Welcome to Aviation SE. A little bit of formatting never hurt anyone though.

You haven't told us the departure airport or the airline so we can't say what regulations applied to the flight, and individual airlines have their own operating procedures too. It's possible that non-weather reasons also applied here (see the answer that @fooot linked to) so I don't know if you'll get a clear answer on this. You might consider asking instead what the applicable regulations are in the country that interests you (USA? Canada?), remembering that an airline's OpSpecs might override them.

Flight was Alaska subsidiary Horizon, origin airport was SEA.  Thanks to 757toga for the info (if I've got it right, PROB40 is ruling and nothing can override that (clear actual condition reports, etc.) though other carriers may have approval for lower mins (hence the "no problem" status of an at-the-same-time Delta flight).   Horizon seems to have operated under US rules:  "EWINS forecast was not considered as an option.  EWINS is authorized in Canada (DSM 5.13) and this may have allowed our flight to operate" (this was the VP Flight Ops quoting the answer he got from the Chief Dispatcher.

Can you post the full METAR/TAF reports? If you can then I might vote to reopen this, but some of the factors in the other question are still relevant. The airline has to consider factors more than just what the TAF says. Legally they never have to cancel a flight if they carry enough holding fuel/enough to divert to a clear airport (unless ATC is over capacity)

@757toga That's a good question. We regularly get questions here that are variations on a theme, e.g. why did my flight fly in circles after taking off? IMO this is a similar case and a general answer that explains the factors in a weather cancellation is more useful than 10 individual analyses that have no wider value. That's why I suggested that the OP focus on the regulations; I don't think we have a good answer on that yet. But, this is all just my opinion and others may well disagree (and vote accordingly), which is totally fine with me.

I've edited the question in response to Ben's request to post the METAR/TAF reports at the end of my questions.  Thanks for all of the answers/comments so far.  They have been helpful as I learn more about aviation weather forecasting, flight rules and such.

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How are the 747-400 rudder pedals and tillers connected to the nose wheel?
The Boeing 747-400 Before Taxi Procedure says (emphasis mine):

[...]
Flight controls .............................................. Check C
[...]
Hold the nose wheel tiller during rudder check to prevent undesired nose wheel movement.
Move the rudder pedals to full travel in both directions and verify:

freedom of movement
that the rudder pedals return to center
correct flight control movement on the EICAS display

The FCOM (14.20.3 Landing Gear - System Description) further says:

Nose Wheel and Body Gear Steering
Nose wheel and body gear steering is powered by hydraulic system 1.
Primary low speed steering control is provided by a nose wheel steering tiller for each pilot. Limited steering control is available through the rudder pedals. The tillers can turn the nose wheels up to 70 degrees in either direction. A pointer on the tiller assembly shows the tiller position relative to the neutral settings. The rudder pedals can be used to turn the nose wheels up to 7 degrees in either direction. Tiller inputs override rudder pedal inputs.

I wonder how the rudder pedals and the tillers are connected. How can the tiller override the pedal input? I assume the tillers would move when the pedals are used, but do the pedals also move when the tiller is rotated?

The easy (non-technical) way to understand 74/75/76/77 steering is that the tiller controls the nose-wheel and the pedals control the rudder. And that they are connected by a spring.
So moving the pedals will A. move the rudder, and B. slightly move the tiller which will move the nose-wheel. Moving the tiller will A. move the nose-wheel, and B. slightly move the rudder.

"moving the pedals will A. move the rudder, and B. slightly move the tiller which will move the nose-wheel" This makes sense to me and is also confirmed by the video ymb1 linked. But are you sure it also true the other way round: "Moving the tiller will [...] slightly move the rudder."?

Looking at this video at 03:20 it does not look like the rudder is moving.

Is there possibly a one way clutching or [hydraulic] valving mechanism to prevent the tiller from feeding back to the pedals and rudder?


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If the Quadcopter is a MIMO Nonlinear System, How Can it be Controlled using PID?
I know that it is very common to control quadcopters using PID algorithms. I'm just wondering if it isn't a problem that the system is highly nonlinear as well as having multiple inputs and outputs. Does the model need to be linearized so that PID algorithms can be used?
Thanks in advance!

I'm voting to close this question as off-topic because it’s more associated with programming and not Aviation per se.

@CarloFelicione How is this programming? This is control theory!

As others have pointed out, MIMO and PID are not exclusive. More to the point, a quadcopter is not really what I think of when discussing the complexities with MIMO. The multirotor is more like a bunch of fully decoupled SISO systems running in parallel.

If the Quadcopter is a MIMO Nonlinear System, How Can it be Controlled using PID?

Exactly as you mention later:

Does the model need to be linearized so that PID algorithms can be used?

Yes, that's correct.
To be a little more explicit, you need two things:

linearise the system (usually around a so-called "working point", e.g. cruise conditions)
de-couple the system (because a linearised MIMO is still a MIMO, you need a set of 4 SISOs)

Achieving the first is fairly simple, as I mentioned, you "only" have to select a working point. The second is instead a little more elaborate, and requires making assumptions on the flight conditions that will be encountered, plus it is not always possible, some vehicles have their dynamics so coupled that de-coupling is not an option (and thus simple PIDs tend to be not a viable option if you have to tune them manually).

You can use PID algorithms to control quite complex systems. After all, it's just a feedback system, which in itself is a robust concept. The problem is rather to control them in an optimal manner; or in other words, how to select the best gains.
All classic methods of gain selection require linearisation of the system around a certain point (flight condition), or multiple points with gain scheduling. But nothing prevents you from tinkering the control system randomly until it works.
This latter is not as stupid as it sounds. Various genetic algorithms have been shown to optimise PIDs for complex non-linear/MIMO systems very successfully.


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How would a low-airspeed fixed-wing drone be designed to reduce effects from gusts and wind shear?
In a story I am working on, the characters need a fixed-wing drone that can rendezvous with a blimp, where by "rendezvous" I mean "come within a very short distance at low relative airspeed".  The challenge is that the blimp will typically be flying through powerful gusts and abrupt wind shear—ideally hurricane-like conditions or worse.  The characters have the time and resources to design and build a custom drone for this purpose, and while I realize that I will probably have to stretch credulity at some point, I'd like the drone's design to at least seem plausible.  Hence my question: What would a drone custom-built for such missions look like?
Here's what I found so far (and which I hope is accurate):

Wing loading is the single largest factor in gust response.  If the only concern were dealing with the wind, the drone would have as much wing loading as could be managed.  However, I also need the rendezvous to happen at low (blimp-relative) airspeed.
A long wingspan is a liability in wind shear, increasing the chance for an asymmetric stall and upping the rolling moment even if there is no stall; the drone should probably have its wings tucked tight to its body.
Given the risk for sudden stalls, the wings should be designed with substantial washout.  They will probably not be swept (in either direction), both because drag reduction is not a driving consideration and because swept wings would tend to push the drone nose-up in a stall.
Anhedral wings, while less good for stability, would cause the drone to bank into crosswinds, which could be useful given the feedback delay the drone pilot has to deal with, and some of the stability could be made up by placing the wings high, creating a pendulum effect.  But maybe stability is the more important factor here.
An inverted tail will cause the drone to both roll and yaw into crosswinds.  I am not sure whether an inverted T or inverted V would make more sense.

I have also found some planes that are intentionally flown into hurricanes: the Lockheed WC-130, the Lockheed WP-3D Orion, the Gulfstream IV, the Douglas DC-8, NASA's ER-2 (a Lockheed U-2 derivative), and the AAI Aerosonde, the only drone on my list.  The last, at least, backs up some of the ideas I have above.

If you remove requirement for fixed-wing i would suggest you to try blimp drone.

@vasin1987 I agree that that would make a lot more sense, but it would also require reworking the plot, so I wanted to at least see if I had any chance of hand-waving something fixed-wing.

The blimp would have far more problems under those conditions.  But a great source of design would be seagulls, as they have a much lower weight to surface area ratio than a/c.  A seagulls in 30 mph winds would be an excellent model.  They have anhedralled wing tips to reduce roll in side gusts, and finely  "brain computer" controlled adjustments to hold them steady in flight.

First, kudos for the well-researched question! You certainly did your homework before coming here.
As you realized yourself, your requirement is quite contradictory. I also wonder what blimp would be flown into hurricanes, but I leave that aside for the moment.
In order to reduce gust susceptibility, a variable configuration would be one option. However, just when you want steady flight in gusty conditions, you also will fly slowly, so variable wing sweep or elaborate wing flaps will not help. The next option which combines low flight speed and high wing loading would be some direct lift, like in the VTOL-designs of the 1960s. Since the direct lift is needed for a short time and when propulsive demands are low, an engine like the Rolls-Royce Pegasus would allow to add lift that is less affected by gusts than the wing.
Make sure your design will draw ample bleed air in order to feed the reaction control system, and maybe a gyro platform to control this system, and the contradiction can be resolved.
Stalls should now be less of a concern, too. Much of how much lift is lost in a stall depends on the choice of airfoils, and if you use one with benign stall characteristics, stalls can be tolerated.
Anhedral is a poor idea, however, especially if you use unswept wings. Only if you have a high wing on a large fuselage can anhedral be justified. Examples are the Harrier jump jet and the F-104 Starfighter, where the large T-tail made wing anhedral necessary. You want moderate dihedral in order to make the aircraft roll in a sideslip. This rolling motion will tilt the lift vector sideways and pull the aircraft away from the sideslip condition.
Also, please reconsider whether there really is such a thing as a pendulum effect.
Regarding the tail: The best one is as symmetrical as practical.

Dear Peter:  big difference between anhedralled wing tips and net anhedral.  Also when one has much larger surface area to weight as in bird vs 747, obviously aerodynamic effects will be more dominant.  Look at a seagull.  Wings are dihedral/anhedral.  Very effective in controlling roll with side gust. Similar to "Lippisch Ears".  As far as pendulum (when NOT turning), yes high wings have more self righting tendency.  But as this is enjoyably academic and interesting, much is learned from all opinions.

Do agree with symmetrical tail.  Perhaps a modern Taube might not be so bad.

@RobertDiGiovanni: Great, you know what Lippisch Ears are! But then you must know that they were added to the He-162 as a kludge when it was too late to reduce wing anhedral. So I wonder what the big difference between them and wing anhedral is. And you are sure about the better self-righting tendencies of high wings? I don't think they make much of a difference.

Witnessed a 10 page discussion of this on another thread, so, I will say look at seagulls.  As far as high wings, it is actually a "double" pendulum. Any displacement of Clift and CG from sky/ground orientation creates a righting moment to restore it not unlike the ballast of a sailing ship.  Lift goes up, weight goes down.  Combine that with balance of area above and below CG from side forces (to prevent rolling from side force) you have stable high wing trainer.  Anhedralling is also seen in horizontal stabilizers to improve turn performance.

No big difference between kludge and anhedral.  We know dipping wing helps in cross wind.

We mix and match as needed.

And let me please compliment you on the incredible amount of info you have provided on this site. I have learned so much from you and other writers.  I would wish to discuss "pendulum" or self righting effect, and another critical parameter, weight to wing area.  I don't think do a 747 slips much, it simply rolls back to level.  Paper airplane, more likely slip/correct.  But this is what makes (gravi)aerodynamics well, better than rocket science.  Let's get a little more tail on that 162!

Consider the "freewing" concept -- each wing is free to pivot, pitch-wise -- as described here https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19700014919.pdf -- google "freewing gust alleviation" for much more-- unfortunately "freewing" is now also the name of a popular manufacturer of radio-controlled model airplanes, so that will complicate your search.
The flexibla nature of the wing structure used in hang gliders and "trikes" (powered hang gliders with wheels) also has some natural load-shedding characteristics.  In addition to the properties created by flexibility, these aircraft can also be considered "freewings" in a way, although the left and right wing panels do not move independently in all but a very few cases.
(A little late now-- is the book finished yet?)


text:
Can a protective mesh protect engines from bird strikes?
Can a protective mesh be installed on the front of the engines to stop birds from entering the engine and causing damage?


text:
Between a helicopter and an airplane, which requires more power to fly, for a given TOW?
For the given same mass (say it 400kgs), which one requires more power (in HP) between helicopter and airplane? Say that both are designed maximum (the required material are considered during design) to minimize the required power.
Edit: Question should be like this: Which one is requires more power to lift a helicopter (to make it hover) and to make a fixed wing airplane fly? Not to make them move at the same speed.

Actually it is flying at the same weight and speed.  Please see answer below.

Either you compare lifting and moving, or you compare the airplane nose up and hovering.  The helicopter requires LESS power to hover.  Why? (compare propeller dimensions and RPM to helicopter rotor).

Airplanes fly by creating lift from their wings. This causes some drag, but good wings have lift/drag ratio's in the range 15-20. That means the lift-dependent drag is only 5% of the lift. Helicopters on the other hand generate lift directly from trust; there is no multiplier involved.
E.g. a 4000 kg plane will have a weight of 40.000 Newton, so the drag incurred will be 2000-3000 Newton. A 4000 kg helicopter will need to produce 40.000 Newton of lift just to hover.
Of course, planes and helicopters both have additional drag from forwards airspeed, and for planes this is obviously unavoidable to prevent stalls.

Dear @mstalters, is it possible you make calculation so will be clearer? A helicopter is lifting the load vertically, while airplane is lifting the load by dragging.

Please see comments below.  Thanks for the discussion.

@AirCraftLover: Added.

Don't forget that you can have arbitrarily large thrust with arbitrarily little power given sufficiently large rotor (at standstill, like the helicopter hover; in motion you need at least thrust times speed, the power that goes to the vehicle)!

Let's look at this in an extremely simplified fashion.
An aircraft with mass $m_{ac}$ stays up in the air by pushing air downwards, or specifically, by giving a mass flow $\dot{m}_A$ [kg/s] of air a certain speed downwards $v_A$ [m/s]. This gives a momentum 'flow' $\dot{m}v$ [kg m/s²] which is the lift force $F_{lift}$ [N]
$$F_g = F_{lift}$$
$$m_{ac}g = \dot{m}_A\cdot v_A$$
The power required for this comes from having to give the air flow a kinetic energy flow $$P_{lift}=\dot{m}_A{v_A}^2$$
This is purely the power required for lift generation (power required to overcome induced drag, specifically). One can see that by making $\dot{m}_A$ arbitrarily large and $v_A$ arbitrarily small (while keeping their product constant), the power requirement can be made arbitrarily small. This can for example be done by making the wings or rotors longer so that they affect a larger air volume (and thus air mass), or by flying faster (so they move through more air, again increasing the mass flow).
However, this assumes perfect efficiency. In reality, wings will experience drag even if no lift is being generated, and the same goes for the fuselage. You will often find a minimum of total power required at some speed such that the induced drag is quite small but the friction drag is not quite as large. This goes for both fixed and rotary wing aircraft. These factors are a result of the aircraft's practical design, not of theoretical considerations.
So, there is no theoretical answer to this question. There is only a practical answer, which is that hovering in a helicopter is very inefficient and requires a lot of power (because it can only affect a small mass of air since it is not allowed to move), so given the constraints in your question (a hovering helo vs a fixed wing at 100kts), the fixed wing is probably more efficient in practice.

We agree on F=ma and vectors.  What is interesting is that with equally opposing forces there is no movement.  The rocket on the test stand is producing Power, the test stand (hopefully) is strong enough to hold it.  Power equation = ForcexSpeed may be correct with the caveat "with no opposing force". In reality, friction, drag or any opposing force can bring velocity to steady state.

And BTW, the solution is to compare the Hughes 269 with the Cessna 150 like @John K and you will see the the difference.  A Cessna 150 flies with around 1/4 of its weight in thrust, the helicopter requires 100% of its weight in thrust but the (higher aspect) rotor is more EFFICIENT than the prop, ok?

@RobertDiGiovanni In steady state, the engine still does work, but the drag does equal work but opposite in sign. Anyway, if we can agree that power is not the same as thrust (which we apparently cannot, despite the discrepancy in units), we can see that the Hughes comes with a 134kW engine for a loaded weight of 703kg, while the Cessna comes with a 75kW engine with a loaded weight of 726kg (numbers from Wikipedia)

Getting somewhere here.  So the helo generates 4x the thrust with 2x the power.  Now let's put a very large prop on an airplane (V-22 Osprey) ?

@RobertDiGiovanni Indeed, a helo is more efficient at converting power to thrust, thanks to the large blades. An airplane can get away with a less efficient (less large, so smaller landing gear required) propeller because it has the highly efficient wings. You'll see that early airplanes had almost comically large propellers since the engines were so small; to this day, you see that turboprops tend to be high-wing (e.g. ATR-72, Dash-8) to accommodate larger props.

From your explanation, you said that v is the velocity of the airflow downward. That is we don't know. So it seems, IMHO, not the answer. That is true that the force required to lift is must be bigger to make the helicopter lift, and than to maintain the constant force as big as the weight (m*g) during hover. But however, the answer of this question is not a simple thing.

As I said, we consider the design is maximum (or optimized in the terminology of @john-k). So, we have to consider the length of the rotor blade AoA, and the length. The length itself can not be too long and too short. If it is too short, required high speed, which at the end will produce induce drag without giving more force if the tips speed exceed the subsonic speed. But if it is to long, it will produce tangential force. So, what is the maximum size?

Rotor and propeller turn, is directly related to the power of engine (we assume that the crank shaft is direct connection or just through bevel gear). But to increase rotor or propeller rotation 10%, will increase the required power 33.1%, plus the lost power absorbed by the air (the efficiency factor). Thereby, we have to maintain the rotor/propeller rotation not so fast. Resize the propeller/rotor until we get the optimum size will probably the best solution.

@AirCraftLover We don't know $v$, but the point is that we can, in theory, make it arbitrarily small by increasing the rotor or wing size. You are asking about a maximum or optimum, but like I said, if we can increase $\dot{m}$ without bound, the power requirement goes to zero. By only looking at the power requirement, we can't find an optimum. Real aircraft are a balance between the cost of adding power (bigger engine) and the cost of reducing power requirement (longer wings).

@AirCraftLover Small nitpick: "produce induce drag without giving more force if the tips speed exceed the subsonic speed" Induced drag is solely the force associated with pushing the air down. The drag due to wings going supersonic falls under parasitic drag

Helicopter rotors need to provide enough thrust to compensate the weight of the helicopter: $T_H = W$.
Fixed wing aeroplanes need to provide enough thrust to overcome drag, while the wing lift compensates for weight. As @MSalters correctly states, the wing provides much more lift than drag, plus there is a fuselage & tail as well.

Torenbeek Synthesis of Subsonic Airplane Design gives some L/W ratios of complete aeroplanes, a medium size turboprop such as the F-27 is listed as having an L/D of 13.8 during takeoff, since aspect ratio A = 12. So this fixed wing aeroplane would have to provide 1/14 of the thrust of a helicopter of the same T/O weight: $T_F = W/14$
Simple impulse theory gives the following relation between thrust T and power P:
$$ T = C_T \cdot \rho A {(\Omega R)}^2$$
$$ P = C_P \cdot \rho A {(\Omega R)}^3$$
 And therefore at constant disk area and tip speed:
$$C_P = \frac{{C_T}^{3/2}}{\sqrt{2}} => P_F = (P_H/14)^{3/2} = 0.1$$
Tip speeds $(\Omega R)$ of propellers and helicopter rotors are comparable, disk area of a fixed wing is lower. So the required power at constant weight for a fixed wing is at least an order of magnitude lower than for a helicopter.

Order of magnitude? That does not match reality at all. E.g. Pilatus PC-12 has 890 kW installed power for 4 740 kg MTOW and comparable size Eurocopter EC155 lifts off at its MTOW 4920 kg with just 1394 kW installed power. That is not even twice as much, long stretch from an order of magnitude!

Comparing thrust of airplane to “thrust” meaning lift of helicopter is totally irrelevant. The larger rotor can produce much more lift for power the same way the aircraft wings do.

@JanHudec How is lift defined in the helicopter fuselage reference frame.

Lift is defined as perpendicular to relative wind (except for helicopter in hover, which has no relative wind; it should be defined as the force opposing weight for continuity).

… and note that the point of defining lift this way is that lift does not do any work on the aircraft. So you only need the induced power and induced power can be reduced by improving efficiency. In contrast to thrust which does work on the aircraft and that sets the minimum engine power required. That's how rotor can produce much more lift than a propeller can produce thrust in forward flight with the same engine power.

When the helicopter takes off it has a vertical velocity, the rotor provides more lift than weight, work is done on the aircraft. It makes no sense to define a vertical force first as thrust and then as lift, depending on direction of the speed vector.

@JanHudec Also, the PC-12 cruise speed is 528 km/h, Never Exceed Speed of the EC155 is 324 km/h. As you know, drag force is quadratically proportional to airspeed. When comparing fixed wing and helicopter installed power, please do so for equivalent velocities.

Parasite drag is quadratically proportional to indicated airspeed. PC-12 can't get even close to 528 km/h indicated. It can get a bit higher than the Vne of the helicopter, but the helicopter isn't drag-limited anyway. The helicopter is limited by retreating blade stall. Not that PC-12 was power-limited either. Unfortunately they don't publish as detailed performance informations to be able to compare the accelerations.

When helicopter takes off vertically, the vertical force is indeed doing work on the aircraft (and can be therefore considered thrust), but note that the velocity is very low, and so is the power.

If you mean for a 400kg helicopter and 400kg fixed wing airplane to move at 100kt, it's generally going to be the helicopter that requires more power since the whole egg beating thrashing mess is a lot less efficient at converting energy to forward speed.
Of course you can make the airplane draggy enough that it can require more power than the helicopter to go 100kt if you want, and there are plenty of those, but I assume we're talking about optimized craft here.

I mean, which one is requires more power to lift a helicopter and to make a fixed wing airplane fly? I agree with you, if the intention is to make it move the same speed, of course helicopter requires more power.

Well, a Hughes 269 weighs 1550 lbs and has 180 hp and is not exactly a vertical rocket with that, and a Cessna 150 weighs 1500 lbs and flies just fine on 100 hp, so there you go.

the helicopter's "wings" (its main rotor blades) are moving through the air at something like 400 MPH at their tips even while the helo itself is sitting in ground effect and not traveling through the air at all. This requires work, and to fly forward, the engine has to overcome the main rotor drag at the same time it has to overcome the fuselage drag. So: more drag for the helo, less for the Cessna 150.

With one caveat: helicopter requires less power when flying forward than when hovering.

General rule of thumb with heavier than air craft is the bigger, slower lifting surface with less interference is more EFFICIENT at converting available power into lifting and/or propelling force.
The helicopter blade has greater efficiency in creating thrust compared with a propeller (see V-22 Osprey),  but the wing is far more efficient at creating lift,  therefore the Cessna 150 requires less POWER to create the same amount of total FORCE (lift and thrust).
You will notice this works for monoplane vs biplane (less interference),  propeller vs fan (much less interference), hi aspect glider vs low aspect speed plane (more efficient wing),  2 bladed helicopter vs 8 bladed helicopter (less interference), bird vs airplane (slower "propeller")
Thrust Force = Power x Efficiency
You will notice birds propel and lift more like helicopters.  Yet they are MORE efficient than the Cessna because their PROPELLERS are proportionally larger and slower.  But we do not see 400 kg birds or 800,000 lb Cessnas because NET thrust is what actually moves the aircraft.
A less EFFICIENT jet has much more power, so much more thrust is available, albeit with more fuel consumption per pound of thrust produced.
Finally, force required to "make the aircraft hover"  and "make the air plane fly"  really is comparable!  One must simply realize in both cases it is FORCE vs DRAG.  So, for direct comparison, they both need to be at flying speed.
POST EDIT - RESPONSE TO COMMENTS
For the benefit of @ MSalters and @ AirCraft Lover an examination of units is in order
F = ma kg m/seconds squared
Now the equation for Power = F x velocity = kg m/seconds squared x m/second
This is an unfortunate corruption of James Watts original weight lifting work with horses (origin of "horsepower") and should read:
Power = massxgravity + massxacceleration to reach speed and Power = massxgravity at constant speed F=ma=weight being lifted. Force vectors are ADDITIVE. Air drag and pulley friction are negligible. (Watt knew to take only the first meter)
Notice distance has NOTHING to do with POWER! (I'm sure football players struggling at the line of scrimmage (Thrust vs Drag Superbowl) appreciate this).
Let's fast forward to the THRUST curve of an Estes C6-0 model rocket and viola! That is what James Watt's horse is doing!
So we can think in terms of thrust, lift, drag, and gravity as ADDITIVE forces (much easier), and better yet, break them down into vertical and horizontal components!

That's some handwaving physics there. Force x Speed = Power.

@MSalters Power opposes drag.  Efficiency produced is really thrust to drag (or lift to drag) ratio.  Could rearrange to thrust or lift force/power=Efficiency.  No hand waving here.  It is the definition of power that includes speed (as d/t) that may need examination.  It takes power (as force/time) to hold up an object (against gravity/time) even if there is no motion.  I have written about James Watt and horsepower and concluded P= weight x d/t should be weight x v/t to accelerate weight to constant speed and simply  P=weight at constant speed for horse.  I wonder what the horse thinks.

I was being friendly there. The equation doesn't even agree in its dimensions. Force is in Newtons, Power in Joules per second, and efficiency is a dimensionless fraction. And no, it doesn't take power to hold up objects against gravity. Unpowered tables do it al the time.

Dear @Robert-DiGivani, is it possible you make to a real calculation? Say, what would be minimum blades and power required to lift a helicopter vertically, and so does with an airplane. From there, that would be clear which one is requires more power.

Dear @Robert-DiGivani, from your equation, force unit is Newton (N), and power unit is Newton-meter/second (N.m/s). Almost there, but however still need correction.

This is where I KISS.  Just balance forces.  For a piston engine the AVERAGE thrust of the pistons = drag of the prop or rotor.  The "lift" force is a product of any airfoil, be it prop, rotor, wing, or noodle.  The efficiency is the LIFT/DRAG ratio.  Just forces, NOT distance!  Notice with Thrust/second = Drag/second time cancels out.  It then becomes an examination of airfoils.  And it is empirical for heavier than air craft.  A Chemist by trade I, not Physicist, but I'm working on it.

Even simpler is thrust per pound of fuel burned over any unit of time.  Remember, a HELICOPTER creates thrust upwards by rotating its wing.  A PLANE uses thrust for forward motion, creating airflow over the wing. In ALL cases, drag must be overcome to produce necessary force.  You will notice the Cessna needs 1/4 of its weight in thrust to fly, the helicopter needs 100 percent, BUT the rotor is more EFFICIENT than the prop!  I'll leave the MATH to you.


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Why does the Biot-Savart law give the induced velocity of a infinitesimal segment of a vortex filament at point P?
My question is about the proof of how the Biot-Savart law can be used for vortex filaments. This is what I have in my textbook:

However, I'm not certain how it was derived - I understand that the Biot-Savart law originally came from electromagnetic theory, but could anyone present the proof for it's use for vortex filaments?
Intuitively though, something seems off - a vortex filament is just made out of lots of individual vortexes put together in a filament, so shouldn't the induced velocity at a point P of an infinitesimal section of the filament be zero unless $d\vec{l}$ and $\vec{r}$ are perpendicular... but clearly it's not the case,
$$
 d\vec{V} = \frac{ d\vec{l} \times  \vec{r} }{ \lvert \vec{r} \lvert ^3}
$$
, which is nonzero even when $d\vec{l}$ and $\vec{r}$ aren't perpendicular

I think this might be a better fit for physics.stackexchange. The vortices may be related to aviation, but proving a theorem is better suited for physics.

The whoda whatzit law for vortex what-the-hecks?  Over my head...

Nevermind guys, I finally found a proof for this, after a long time of googling:
https://www.whoi.edu/fileserver.do?id=218347&pt=10&p=116694
Basically, you use the differential definition of vorticity (w=curl of velocity) and the incompressibility condition (divergence of velocity = 0), then combine them and solve.

Can you please summarize the results into your answer? Otherwise, it defies the purpose of SE.

Why would

the induced velocity at a point P of an infinitesimal section of the filament be zero unless $d\vec{l}$ and $\vec{r}$ are perpendicular

?
The cross product in
$$
 \frac{ d\vec{l} \times  \vec{r} }{ \lvert \vec{r} \lvert ^3}
$$
does not change signs when going over the $\frac{\pi}{2}$ angle in between vectors, so the infinitesimal filament sections $d\vec{l^-}$ (immediately before $d\vec{l}$) and $d\vec{l^+}$ (immediately after $d\vec{l}$) have contributions to $d\vec{V}$ that do not cancel each other out. Therefore, $\vec{V}$ is not just a function of a single filament section, but of the entire filament, hence the integral.
If you check the next section in the source of your image, titled Straight-vortex case you will see this in action.

Nevermind, I found a proof for it:
https://www.whoi.edu/fileserver.do?id=218347&pt=10&p=116694


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Why were the ailerons of the Fokker Dr. I (Red Baron) located on the top wing?

Photos' source are here: Fokker Dr. I (Red Baron), and here: a Sopwith triplane. They are probably not the actual airplane but only (claimed) authentic models.
It makes sense to me that the Sopwith had ailerons on all three wings. But the Dr. I only had them on the top wing.
My question is, what is the aerodynamic explanation of this design? Why not have ailerons on the middle and on the bottom wing? Which design is better?

The top wing is always the most efficient, as the lower ones suffer reduced lift due to interference from the air flowing beneath the one above. Ailerons will therefore be most effective on the top wing. If they do the job OK then there is no need for any more.
Reasons for adding more below include:

The upper set are likely to be (or found to be) inadequate.
Making both wings identical to simplify production.

The outermost part of the wing is also the most effective position. So if a short aileron is inadequate then the designer faces a choice of two less efficient additions, to widen it inboard or to place another one underneath. Another issue is which brings more drag or weight penalties, a small aileron which must be sharply deflected or a larger aileron which needs more subtle movement.
Such decisions were often based on the designer's personal experience and preference rather than firm knowledge. They would stick with what they knew rather than take arbitrary risks. Historically the Sopwith came first, following their earlier successful biplanes such as the 1 1/2 Strutter and Pup in having identical upper and lower wings with short ailerons. Fokker responded in his own equally successful house style, with a single aileron on the top wing only, having a long span and outboard balance tab.
Which design is better depends on how you define "better" - less weight, less drag during turns, less cost, optimal distribution of aerodynamic forces through the structure? The basic technical tradeoff is between a single large and strong upper aileron vs. several smaller and lightly-built lower ones. I doubt if anybody has ever crunched realistic numbers.

The upper wing of the Fokker Dr.I is larger in span and chord than the other two. Adding ailerons on the lower wings would had

added less effect for additional mechanical complexity,
and on the lower wing they would be subject to damage upon ground contact.

I know that it is hard to argue here that ailerons must be protected from ground contact, but this was a real risk in the rough operational circumstances of World War I. Pilots would be assigned to operational units with 20 hours of total flight experience and the light and small airplanes were at the mercy of sudden gusts or bumps in the airfields.
Why else would the Dr.I have skids below the lower wingtips?

I circled the wing tip skid in red (picture source).

That also make sense that the upper wing is larger and longer.


text:
is it possible that setting a lower airspeed makes a steeper climb?
I'm now studying a Flight Management System (FMS), and it says that when the FMC predicts that the airplane will not reach an ALT constraint, the CDU will display "UNABLE NEXT ALT" and then we have to manually set a lower airspeed to provide a steeper climb.
But, I've learned that when we pitch up, the aircraft slows down because of more drag.
I also remember that when an aircraft increases speed, pitch goes up because of the increase in thrust and more lift from the wings.
So why is a decrease in speed needed to make the FMC altitude constraint, rather than an increase?

Do you fly this airplane with a FMS?

When you choose a slower airspeed, aren't you choosing a speed that is closer to Vx?  Are you familiar with what Vx means?

I think you have it backwards, it is not that a slower airspeed produces a steeper climb, it it that a steeper climb, given the same power, will cause a slower airspeed.
If the only constraint you're setting the FMS is the airspeed to maintain for a constant thrust, then it will pitch up more, increasing the rate of climb, which will slow the airspeed.

Initially, pitching up will trade airspeed for altitude. Going from a climb at 300 knots to a climb at 250 knots means pulling back & increasing the rate of climb while the speed decreases.
For a close-in fix to the climb profile, this is often sufficient to rapidly gain several hundred feet above the FMC prediction and solve the situation.
For a steady climb, the slower speed is typically a steeper sustained angle, mainly because the parasite drag goes up with airspeed (and at typical climb speed ranges, parasite drag is the larger contributor to total drag than induced drag is). So for a case when you're many miles from the restriction, the improved climb angle at the lower speed will help you meet the constraint. (Note that this only works to a point, getting slower than the max angle of climb speed, other things come into play & you're then hurting your climb performance. But that Max Angle speed is typically well below normal climb speeds.)
Finally, as a matter of planning, it takes time -- and therefore distance -- to accelerate.  Delaying that acceleration until past the constraint can be helpful too. There is a departure out of KBWI with an altitude constraint at or above 11,000' that can be hard to make under certain conditions if  you let the FMC accelerate the aircraft from 250 (speed below 10,000') to a fast climb speed (say, 300 knots) before getting very far above 10. The easy solution is to keep the speed at 250 until through 11,000, which makes the restriction easily, and then accelerate.

If you keep a constant power setting then the way to decrease your speed is to pitch up. Simply, this will increase your rate of climb as more of your total vector is ‘upward’ rather than ‘forward’. The CDU is assuming a constant power setting during the climb so will naturally calculate a higher rate of climb for a lower airspeed.


text:
Is it possible to Fly Aircraft on Mars?
I have been fascinated with the thought of flying all kind of aircraft on Mars. To begin, I don't know much about the fundamental things or requistes for flying. Is it possible with the the things we currently have? What kind of infrastructure would be needed to fly there? Assuming GPS navigation is already set up there. What would be the design of the aircraft and the runways? Would it be like we have on Earth or completely revolutionised Idea? Kindly not mention the costs of flying there.

Theoretically, yes! At least that's what NASA is planning on doing next year

https://mars.nasa.gov/technology/helicopter/

Related: What kind of wing-surface area would Curiosity need to be able to fly in the atmosphere of Mars?

highly related: XKCD what if #30: Interplanetary Cessna. It even answers for other planets (but only for one type of aircraft)

There is many questions in your question. each could be a separate question, some being good one for space.stackexchange. E.g. GPS is not needed to navigate (VOR, DME, and other ground-based navigation aids are older and highly reliable), designing a runway take into account many parameters such as available materials (planet-specific) and desired strength (depends on the aircraft you want), ...

@Manu H: Indeed, some of us here probably learned to fly before GPS even existed.  I certainly learned long before it was generally available to consumers, and still seldom use it for anything, preferring to navigate by visual references.

I think the fundamental issue you'd face on Mars is the lack of an atmosphere.  Atmospheric pressure on Mars is less than 1% of that on Earth.  Most of our machines for flight on this planet rely on atmospheric pressure under the wings or rotors which wouldn't work on Mars.
The other issue we'd have to overcome would be a fuel source.  Petrolium based fuels such as Gasoline, Diesel and Kerosine in both piston and jet engines again rely on a certain amount of oxygen pulled in from the atmosphere.  Take this away and you'd be looking to replace the oxygen element of fuel with some type of liquid oxygen.
Maintaining a habitable cabin pressure may also be difficult which may mean it was easier to have flight crew and passengers wear "space suits" like those worn by Blackbird pilots.
It's fair to say, the only real kind of flying machine we have tried and tested here on Earth that would work on Mars is a rocket.  Launching would be fairly simple as you can launch from a pad.  Therefore you'd be looking at launchpads rather than runways.  You'd also had a problem landing as most rockets either splash down in an ocean or land on a landing drone ship in the ocean.  Obviously there are no liquid oceans as you'd be looking at landing on the surface.  You would also find you couldn't glide in like the shuttle used to as the really depends on an atmosphere.
I think navigation and traffic control would be the least of your worries but if you could find a craft that could take of and land predictably, you'd obviously need to implement some sort of airspace control network.  Equipment such as radios and radar would likely work much as they do on this planet.

NASA has just a couple of weeks ago launched a helicopter to Mars. It's only a test mission, but they seem to think that rockets are not the only thing that works. Also, I don't get the problem with the drone ships. The only reason SpaceX lands some of their rockets on drone ships is that there is no land where they are landing. The lack of an ocean wouldn't be a problem in this case, it would be a benefit. There is no need to have a drone ship with complex computers that allow it to hold position with centimeter accuracy, if you could just land on land.

SpaceX doesn't need an ocean because of the drone ships, they need the drone ships because of the ocean.

A drone ship on the ocean will no doubt provide more “give” in the landing platform than solid rock.  Not insurmountable I agree.  I suspect the NASA helicopter was nothing like anything many of the pilots here fly.

SpaceX does very much prefer to land their rockets on land. They only use the drone ships when they don't have enough fuel for the boost back burn and can't get back to land. The landing on the drone ship is much, much harder because you have to make sure your landing pad stays put. That's not a problem you have on land. Another problem they have is the vertical movement of the drone ship.

All the successful Mars landers to date have landed on the surface, using rockets (sometimes in conjunction with parachutes & airbags): https://www.space.com/33483-mars-landing-technology-viking-curiosity-rover.html  FWIW, I fly around much of the western US, only rarely interacting with any sort of airspace control network.

All the failed ones too, a good landing is survivable and a great one is one where you can reuse the aircraft.  I’m not saying landing a rocket on solid rock is impossible, I’m just saying the success rate isn’t as good as landing on a runway.

… and the reason launches are done over the ocean is that there is very little the rocket could destroy if it fails and crashes there, while most land is inhabited. Baikonur is in the middle of sparsely populated steppe where both launches and landings can be done on land without such risk, but there isn't as much uninhabited land in America, so all the launches are from the shore towards the ocean.

@Jan Hudec: Actually the choice of KSC as the US's primary launch site was as much because it's about as far south (closer to the equator) as you can get in the continental US.  The closer you are to the equator, the easier it is to launch into equatorial orbits.

@jamesqf, being close to the equator is important, but having open area in the launch direction is also a factor. Note that the KSC launch site is not at the southernmost tip of Florida, where there are other islands around, but further north along the eastern coast, where it has nothing but open sea to the east. Also launches to polar orbits are not done from there, as they'd pass over the cities on the coast, but from Vandenberg, which has open sea to the south.

you're in luck, somebody did the calculation/ simulation.

Mars: To see what happens to aircraft on Mars, we turn to X-Plane.

X-Plane is the most advanced flight simulator in the world. The product of 20 years of obsessive labor by a hardcore aeronautics enthusiast who uses capslock a lot when talking about planes, it actually simulates the flow of air over every piece of an aircraft’s body as it flies. This makes it a valuable research tool, since it can accurately simulate entirely new aircraft designs—and new environments.

In particular, if you change the X-Plane config file to reduce gravity, thin the atmosphere, and shrink the radius of the planet it can simulate flight on Mars. (Note: Thank you to Tom J and the folks in the X-Plane community for their help with aerodynamic calculations in different atmospheres.)

X-Plane tells us that flight on Mars is difficult, but not impossible. NASA knows this, and has considered surveying Mars by airplane. The tricky thing is that with so little atmosphere, to get any lift, you have to go fast. You need to approach Mach 1 just to get off the ground, and once you get moving, you have so much inertia that it’s hard to change course—if you turn, your plane rotates, but keeps moving in the original direction. The X-Plane author compared piloting Martian aircraft to flying a supersonic ocean liner.

Our Cessna 172 isn’t up to the challenge. Launched from 1 km, it doesn’t build up enough speed to pull out of a dive, and plows into the Martian terrain at over 60 m/s (135 mph). If dropped from four or five kilometers, it could gain enough speed to pull up into a glide—at over half the speed of sound. The landing would not be survivable.

so, you would need runways that look at each other, as turning is almost unfeasible, and these runways will have to be really long, as you have to get to Mach 1 to take off and you have to decelerate from that same speed once you touch down.
GPS would not be really needed.


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How has afterburner design changed since its inception?
I came across a NACA summary paper on the past 5 years of afterburner development and started reading cause it seemed interesting. However, a couple of pages in I scrolled back up and realized that it was written in 1956. How much have afterburners changed since then and would the information contained in the article still be relevant?

This article suggests that they have - alluding to improvements in fuel diffusion... adding additional igniters downstream of the first to even out the burn... vectoring improvements to better keep the superheated gasses in the center and avoid overheating the structure at the sides... introducing cool air in a ring around the superheated exhaust stream to aid in a smaller thermal footprint... incorporation of stealth materials... to name a few: https://www.airspacemag.com/flight-today/how-things-work-afterburners-18481403/?page=1

@AaronHolmes put your answer down there in the Answer box.

@FreeMan: I don’t mind doing that. I just don’t really know a lot about afterburners. I thought I’d let someone with more experience have a crack at it.

This article suggests that they have. To name a few:

alluding to improvements in fuel diffusion
adding additional igniters downstream of the first to even out the burn
vectoring improvements to better keep the superheated gasses in the center and avoid overheating the structure at the sides
introducing cool air in a ring around the superheated exhaust stream to aid in a smaller thermal footprint
incorporation of stealth materials

Afterburners have changed, that’s true, but their change is not too drastic. The main operation principles and parts still remain the same. So the paper you read from 1950s is still applicable to a large degree. However as should any practicing engineer, you should keep yourself updated by following closely the recent research on them.
One main area of extensive research that is still going on nowadays and might have changed the most along with materials is the combustion efficiency and effective mixing of the fuel and air while keeping short afterburner lengths. Also combustion and flame stability have been important issues for which much research have been carried out.


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Jet Stream effects at 60.000 ft
Today's airplanes cruise at 40,000 feet. Easterly flight times are longer than their westerly components due to the direction of the jet stream. Comercial supersonic flights did/will cruise at 60,000 feet.
Will the force of the jet stream be as strong at this altitude and affect flight times as strongly as it is at today's cruise altitude?

Welcome to Av.SE, Michael!

Direction of the jet streams and prevailing winds depends on latitude, so easterly flights only take longer between +-30 and 60 degrees. The opposite is true closer to the poles and equator. see https://aviation.stackexchange.com/questions/2549/why-do-flight-times-differ-between-traveling-east-versus-traveling-west/88498#88498

What are the altitude of the easterly winds at the poles and equator?

You need to use an "@username" to notify a commenter other than the author of the question or answer if you want a response. The prevailing winds exist at all altitudes from the ground to the jet stream and above. Speeds vary significantly by altitude and weather pattern. If you are asking specifically right at 90deg and 0deg, theoretically the answer is zero as that should be pure vertical motion.

Wikipedia notes that

The strongest jet streams are the polar jets, at 9–12 km (5.6–7.5 mi; 30,000–39,000 ft) above sea level, and the higher altitude and somewhat weaker subtropical jets at 10–16 km (6.2–9.9 mi; 33,000–52,000 ft).

Thus, anything cruising at 60,000' should be over a mile above the highest of these altitudes, and thus wouldn't be expected to experience the same change in groundspeed that aircraft flying lower, and thus in the JetStream, would have.

Source: NOAA's "SkyJinks" website
One more:

"The altitude of the jet stream is a function of the vertical and horizontal distribution of temperature in the earth's atmosphere. You may recall that the earth's atmosphere is broken into several layers, or 'spheres.' The troposphere (in Latin, literally the 'turning' or 'changing' sphere) is the lowest layer of the earth's atmosphere, ranging in depth from around nine kilometers at the poles to around 16 kilometers at the equator. Within the troposphere, the temperature decreases with altitude, at a rate of approximately seven degrees Celsius per kilometer. Starting where the troposphere leaves off, the stratosphere (literally the 'layered' sphere) extends to an altitude of roughly 45 to 50 kilometers. Temperatures within the stratosphere increase with height, a phenomenon known as a temperature inversion. The transition between the troposphere and the stratosphere, called the tropopause, represents the coldest point of the troposphere. It is at this level, just under the tropopause, that the jet stream resides.

Source: Scientific American
Discussing the Concorde, Wikipedia notes that

With no other civil traffic operating at its cruising altitude of about 56,000 ft (17,000 m), Concorde had exclusive use of dedicated oceanic airways, or "tracks", separate from the North Atlantic Tracks, the routes used by other aircraft to cross the Atlantic. Due to the significantly less variable nature of high altitude winds compared to those at standard cruising altitudes, these dedicated SST tracks had fixed co-ordinates, unlike the standard routes at lower altitudes, whose co-ordinates are replotted twice daily based on forecast weather patterns (jetstreams).


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Does the weight of a pilot exponentially affect the necessary weight of man-powered aircraft?
The MIT Daedalus, a man-powered craft, weighed about 31kg. This will be significantly lighter than its pilot during its record setting trip, though I can't confirm his exact weight. Still, this raises the question: If the pilot had been half the weight, would you be able to safely reduce the Daedalus weight by more than half (aka: exponentially instead of linearly)?
Since smaller birds seem to have an advantage over larger ones, I would assume it'd be easier to achieve flight with a smaller pilot.
Note that the pilot is not half weight due to being malnourished or lacking musculature. The half weight was meant as an easy example. If someone has an example for a 20% lighter pilot, that should be useful.

Are you familiar with surface to volume ratio as it applies to scaling?

Is this about getting airborne, or flying the distance between Crete and Santorini? Scaling up lets distances shrink and vice versa.

While the weight of the pilot of the Daedalus 88, Kanellos Kanellopoulos, is unknown/hard to find...I would posit that an adult male 1/2 his weight would lack the musculature to power the thing.
Going by other world class cyclists weight (Greg LeMond, Lance Armstrong, Eddy Merckx) 150-160lbs (68-72kg)....a 35kg human would absolutely lack the power to drive that thing, no matter how little it weighs.
Additionally, the airframe has to be strong enough to handle the thrust and torsion. Not just the weight of the pilot.
I expect if they could have built a 16kg airframe strong enough, they would have.

Shotaro Iribe weighed 60kg, and he was a national champion. Eri Yonamine is 50kg. 12yo Gurdev Thakur managed 45km over hills in under three hours, which is fairly impressive when the record for most distance cycled in one hour is only 55km. So if the cyclist was half the weight due to lack of muscle mass, and not due to smaller general size, this would be correct. But it doesn't actually answer the question.

Is it possible to answer this comparing hill climbing vs flat races? If a smaller human has a power to weight advantage this should show up as a shift in body weight of winners over flat races where presumably power to frontal area/drag is more important.

This would make sense in the aviation world by considering the pilot as an engine.  We would need to know the feasible output of the engine vs the all-up weight of the aircraft.
Weight differential of the pilots would be small enough to limit aircraft modifications to perhaps an extra wing panel.  G limits for these delicate man-powered aircraft are low.
The real difference is how much drag is saved with a lighter pilot vs how much less thrust the smaller pilot can provide.  Rearranging the lift equation shows lift is proportional to Area.

Area = Lift/(density×Coefficient×Velocity$^2$)

Then compare Drag of reduced wing area to thrust output of the smaller pilot.  It may turn out the larger pilot, or engine, will be better.

The engine weight vs power comparison is spot on!

I think a human generates like a quarter a horsepower? We could go with that estimate, and say the half-weight pilot generates 1/8th HP. We could have them weigh 40kg and 80kg.

@Nail really looking at thrust here.  A large slow prop is very efficient, as is a slow aircraft with large wings for lifting.  Interestingly, with a larger pilot, Mass = Pilot + Airplane will have a higher % pilot.  I would find the pounds of drag to fly (based on glide ratio), and consider a Daedalus built for two!


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Can the assumed temperature be lower than the flat rated temperature?
I was playing with reduced takeoff thrust in a Boeing 737-800 simulator yesterday and was surprised that I could enter assumed temperatures below the flat rated temperature of the engines. The permissible range was from OAT up to 70°C.
The FCOM (11.40.58 - Flight Management, Navigation - FMC Preflight) says:

Selected Temperature (SEL)
Entry of an assumed temperature calculates a reduced thrust takeoff N1.
Entry can be made in degrees C or degrees F.
Maximum allowable entry is 70 degrees C (158 degrees F). The FMC, however,
will limit the N1 to 25% takeoff reduction.

The lower limit is never mentioned. But according to the following chart, the flat rated temperature of a CFM56 engine is 30°C for TO thrust:

(source)
I would expect the thrust to be constant for all assumed temperatures below 30°C, but I saw a reduction in the Takeoff N1 field of the FMC when increasing the temperature in this range:

The blue points are what the simulator tells me at a sea level airport in ISA conditions, the red model is calculated with
$$ N1(T) = N1_\mathrm{max} \times \sqrt{\frac{T_0}{T}} $$
where $ T_0 = 288.15 \, \mathrm{K} $ is the OAT (15°C).
Is this a bug in the simulator? Or can the assumed temperature actually be lower?

Related:

What is the difference between Assumed Temperature and Derate takeoff thrust?
What is a flat rated engine?

If an OAT lower than around 30 degrees is added, according to the charts, the thrust is on the flat part of the curve.  The simulator should just be giving you "flat rated thrust" at any temperature below 30.  More interesting is the N1 curve, which peaks at 30.

The performance software for commercial aircraft of a different model I‘ve used in the past a very long time ago did indicate slightly higher thrust with decreasing temperatures below flat rated temperature. This change was not of any significant magnitude though.


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What type of aircraft would fly with an orange light at night?
I know planes usually fly with red lights which seem small in the sky, but last night i saw a bright orange blinking light traveling through the sky; it was bigger then a star but was orange. I thought it might be a helicopter, although I read it could have been a comet or a meteor.

How fast was the blinking? Continuous?

If you can include your latitude & longitude, and the approximate time you saw it then it might be possible to pinpoint what you saw (eg, might have been the ISS - which is trackable)

I'm sorry but it seems inconceivable that you would not be able to distinguish a comet, from a meteor, from something else. This not a well-researched question. Maybe start by asking a question of Astronomy Stack Exchange.  Anyway, was the blinking regular or irregular?

While normally red in colour, anti collision lights can often appear more or less orange, depending on the background and atmospheric conditions. It is impossible to determine the aircraft type just based on the fact that it has a orange light on it, since almost all aircraft have reddish orange anti collision lights.

Needs the one frame when the light is on and reflecting off the bottom of the plane and looking very orange...

Others have mentioned red anti-collision beacons that might look orange in certain environmental conditions, but there's only one kind of aircraft that shows an intermittent orange light: balloons. The flame used to heat the balloon is orange, and most balloons are translucent enough that the whole balloon can appear orange when the flame is lit.
There wouldn't be a blinking orange light, but the flame is often used in short bursts to keep the balloon inflated, so it would be intermittent. Balloons don't move very fast, so if it was really "travelling across the sky" as you describe, perhaps this isn't the answer.
A meteor (which you've already guessed at) moves a lot faster, but tends to glow orange or white for a few seconds and then disappear: I don't think I've ever seen a blinking one.

Hot air balloons don't really fly at night, though.

@J.Hougaard It's rare but permitted. Dawn and dusk flights are a lot more common and the glow is still very visible then.

Not saying it doesn't happen, but I imagine most people try to avoid ending up in a situation where they have to land a hot air balloon in the dark (since you can't really control where you're going...)

Orange isn't a standard color for aircraft lighting. Best guess would be reflected sunlight on a metallic aircraft, although any number of other explanations are at least possible.

Would the sun still reflect on an aircraft at midnight?

Don't believe the OP mentioned a time. Near sunset, such a reflection is possible; after the End of Evening Nautical Twilight, not so much.

@user1871 Also depends on your latitude (It can be daylight at midnight), and the altitude of the object supposedly reflecting the sun.

In this new era of widespread droning, anything is possible.  People can and do put all kinds of non-authorized lighting on drones.  Maybe you saw a drone, possibly flown by an amateur hobbyist in a manner non-compliant with regulations.


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Why can a plane land in zero visibility conditions but not take off in zero visibility?
After waiting hours for a flight departure delayed because of heavy fog, I wonder why planes can perform an instrument landing with no visibility, but cannot take off without a minimum visibility.

They can't land without a minimum visibility either. The reason isn't because they have to see to land, it's because they have to see to taxi.

Using a HUD, an airliner can take off with visibility as low as 300’, and land with visibility as low as 600’. Without a HUD, you need 500’ (and appropriate runway lights/markings, plus regulatory approval) for takeoff, and whatever your autopilot Cat III autoland system (along with the landing runway) is approved to for landing. (Or 1800’ for cases of no Cat II / Cat III capability.)
As noted, a pure “zero visibility” landing (a Cat IIIc) doesn’t yet exist, because while the autopilot can bring the aircraft to a stop on the runway, some visibility (300’) is still required to taxi clear and get to the parking spot.
Thankfully, visibilities below 500-600’ are pretty rare in most places. Not all airports have the taxiway lighting required for low-visibility (typically, below 1200’ lateral visibility) taxi operations. That is an airport limitation, not an aircraft limit.
Talking about so many feet of lateral visibility, the actual nomenclature is Runway Visual Range, or RVR. That is reported in feet in the U.S., and often in meters elsewhere.

Though your question does not explicitly say so, it implies that you're talking about flights that are carrying passengers or cargo for hire, and other answers address that. However, taking your question as explicitly stated, in other words not just air carrier operations, and applying it to U.S. operations, the phrase:

cannot take off without a minimum visibility

is not always correct.
There is no takeoff minimum required for Part 91 operations (private aircraft if you will). In the 2017 edition of https://www.faa.gov/regulations_policies/handbooks_manuals/aviation/instrument_procedures_handbook/media/FAA-H-8083-16B.pdf, you will find on page 1-8 the following:

Aircraft operating under 14 CFR Part 91 are not required to
  comply with established takeoff minimums. Legally, a zero/
  zero departure may be made, but it is never advisable.

Takeoffs by corporate and personal aircraft without established takeoff minimums were common, though not frequent, in the rainy and foggy U.S. Pacific Northwest weather prior to my retirement, and I have no doubt Part 91 operators still take advantage of that regulatory flexibility. I made many such takeoffs without incident as did fellow corporate and personal pilots at my home field.
By the way, while everyone I think understands what they're saying when they say zero visibility, you can always see something, especially if you're sitting down low in a small aircraft.

As part of my training for my instrument rating I practiced zero-zero takeoffs. Basically, line up put the foggles on and take off. I definitely wouldn’t do it for real.

They can take off at very limited visibility, and they do. Key is then of course to make sure that no other aircraft are on the runway, and this is in fact how the deadliest crash in aviation history happened: the collision between two B747s at Tenerife Airport in 1977.
The accident was not caused by the reduced visibility alone, but as usual by a long chain of individual factors. If all proper infrastructure and procedures are in place, take-off can take place in very low visibility.

I have departed in zero-zero conditions (twice.) In both cases, there was another airport very nearby without this problem and I was operating under Part 91.
The biggest danger in a zero-zero takeoff, after unseen planes/objects on the runway, is what are you going to do if the aircraft decides to malfunction on liftoff?  Lose and engine (or the only engine.)
Commercial operations have limits for this, and other, reasons.

Another factor is that if an airplane is on the ground, it can generally remain safely on the ground indefinitely.  If a plane is in the air when bad weather conditions arise, and the plane doesn't have enough fuel to either wait out the bad weather conditions or go someplace the weather is better, the plane is going to land in bad weather conditions.

When an aircraft lands in near zero visibility, it is a fully automated process that is actually done without input from the pilot.  The pilots are only needed to  taxi off the runway and park the aircraft. It is called "Autoland" and many of today's aircraft have it.
As of today no manufacturer has designed an aircraft with "Auto Takeoff" but I suppose it is not impossible.

A HUD can be approved for Cat IIIa operations down to 6/6/3 for landing (i.e. RVR off 600’ at touchdown + midpoint and 300’ at the rollout sensor).

Under Part 91 operations an aircraft can takeoff with zero visibility (however dangerous that might be).

It all depends  upon the equipment the aircraft has board, the infrastructure at the airport that is landing on, and the regulations under which the flight is being conducted under.   Aircraft can, under certain circumstances, land in zero visibility conditions.  But this requires both the aircraft be equipped to fly a Cat IIIc ILS, approach.  The aircrew must have undergone specific training and certification to fly Cat IIIc instrument approaches, and the airport  must be equipped with the infrastructure on the ground to support a Cat IIIc ILS approach.   As for conducting a zero zero take off, for revenue carrying commercial and charter flights conducted under parts 121 and 135, this is not allowed; minimum visibilities are 1 sm with 2 or less engines and 1/2 sm with more than 2 engines.  You may, however,  conduct zero zero departures while operating under Part 91 regulations.
Zero zero instrument landings are flown by specifically designed auto pilot for this purpose.

No aircraft nor airport is certified for Cat IIIc as of now. https://aviation.stackexchange.com/questions/29375/is-it-true-an-ils-cat-iiic-can-automatically-taxi-in-an-aircraft

Does "zero zero" mean zero visibility, zero ceiling? If you have zero visibility, doesn't that automatically imply zero ceiling? Could you please explain this a little? I assume  "1 sm" is 1 statute mile. Thanks in advance!

Part 121 allows takeoffs with considerably less than 1/2 statute mile of visibility, given the right runway markings and/or lighting (even for 2-engine aircraft). I suspect that Part 135 does as well. Those limits are published in the airline OpSpecs, not necessarily on the Jeppesen charts.

@Penguin With visibility approaching zero, the ceiling will probably be zero, although it may be reported using a different METAR term like Indefinite or Sky Obscured or Vertical Visibility 100’ or something similar. Once ceiling gets below 200’  visibility gets below 1000’, everything is predicated on RVR (lateral visibility) anyway. Perfect “zero-zero” is very rare, but yes you’d reach zero ceiling first. And yes, “sm” = statute mile(s).

The minimum takeoff visibilities you mention are the standard minimum visibilities. The actual minimum required takeoff or RVR in any given situation may be substantially less, but RVR from 600 to 1800 are commonly authorized for §121 or §135 air carriers.


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How to calculate the distance between two waypoints on an approach plate?

Can you please show me how to calculate the distance between the 2 waypoints A and C as shown on the picture? I would really appreciate it. I am having difficulties to calculate the distance between 2 waypoints.

Is there any Aviation Forum out there, where I can get some help? Please advise!!!!!!!!!!!!!!! Thanks

Are you having issues with the math, or with reading the plate?


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Why are crop dusting spray booms not integrated into the wing?
All crop dusting aircraft seem to have an externally mounted spray boom for distributing their pesticides, I imagine this significantly increases drag & therefor performance.
What gives? why not integrate the spray system into the wing?
I have read that crop dusters spray booms remain short of the wing tip to prevent interaction with the wingtip vortices, perhaps another aerodynamic interaction along the main span of the wing prevents this integration?
thanks for considering my question!

That would pigeonhole the plane into a spray-only platform.  These planes can tow gliders, carry fire retardant, and fit other types of hoppers to spread different types of material.  Removable spray bars are an industry requirement, not a design oversight.

also looks (but I've not done studies) that placing the nozzles below the flow of air over and under the wing helps in better dispersion of the chemicals, and at least helps avoid contaminating aircraft surfaces with said chemicals.

Maybe it's because the trailing edge of the wing is an aileron or flap? Mounting sprayers on a moving control surface would require flexible connections, and add stress to the hinges and actuators.

I think that the drag is negligible for this activity

Different chemicals use different nozzles and delivery systems.  Many of the spray agents are very corrosive.  Turbulence by the sprayer can be a good thing, as it aids in mixing.
Additionally drag is not a big issue as crop dusters fly slow, and tend to have good power to weight ratios, so the drag of a spray assembly at 70 knots or 60 knots, is not a big concern.  Mixing (with air of) chemicals, reduction of airframe corrosion, and ease of removal are larger concerns.


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What advantage does Kawasaki C-2 have with a twin-engine configuration as compared to its competitors like An-70 or A-400?
While reading about various military transports, I came across the specs of Kawasaki C-2 and a few questions popped up. So please bear with me here if I list them all together as points, I hope someone here will help me understand more by answering this.
1) What advantage does the C-2's get with a twin-engine setup when compared to the 4 engine set up in most of its competitors (A-400, An 70/188) which have similar MTOW's?
2) The next question is related to C-2's MTOW & payload - with a 145tn MTOW the An 70/188 has a max payload of 47tn but with a 141tn MTOW the max payload of C-2 is a full 10tn less at 37tn. Also, considering that the dimensions are very close to each other what is the reason for such a huge difference?
3) Next is again payload & thrust related - the max thrust generated by C-2's twin engines is at 532 kN (266 x 2) its close to the 468-568 kN (117 or 142 x 4) of the IL76MD-90A and higher than the 468-500 kN (117 or 125 x 4) listed for the Y-20, but both these platforms list a much higher payload capability (~50-66 tn max.) while I understand that the MTOW of the other two are much higher which gives them much higher fuel load and range, shouldn't the similar thrust mean it can carry similar payloads even if it is for a shorter range? Can someone explain to me why such a huge divergence?
4) Lastly is a bit general, I couldn't find the max fuel capacity of the C-2 from any open sources or articles, so, if someone could answer that and share any interesting things they know about C-2 that would be great.

I'll leave answer to people who know more but I am pretty sure most differences are due to the customer specifying much higher cruise speed than competing designs have. Certainly this explains your engine questions. The customer also specified a minimum payload which C-2 exceeds by wide margin. Higher payload simply was not needed. Possibility of exporting to other customers with other needs only materialized after C-2 was designed.

Far too many questions for the SE format.

@VilleNiemi Thanks..I was also hoping to hear from someone with some knowledge in this area, hence the post here...

@FreeMan I was trying to just club related questions under one post in order to prevent from being flagged as duplicates, also considering that all are related to one Aircraft. So do you suggest that I should post these as separate questions?